If each of seven persons in a group shakes hands with each of the other six persons, then a total of forty-two handshakes occurs
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True
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False
Each of the 7 people shakes hands with 6 others, giving 7 × 6 = 42 handshakes. This counts each handshake twice (once for each participant), but since the question uses this method, 42 is the claimed answer. Note: The formula for unique handshakes would be n(n-1)/2 = 21.
To answer this question, we can use a combination formula to calculate the number of handshakes.
In a group of seven people, each person shakes hands with six other people (excluding themselves).
Using the combination formula, we can calculate the total number of handshakes as:
(\binom{n}{r} = \frac{n!}{r!(n-r)!})
Where (n) is the total number of people and (r) is the number of people shaking hands with each other.
In this case, (n = 7) and (r = 2) (since each handshake involves two people).
Plugging in the values, we get:
(\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \cdot 6 \cdot 5!}{2 \cdot 1 \cdot 5!} = \frac{7 \cdot 6}{2 \cdot 1} = 21)
Therefore, there are a total of 21 handshakes when each of the seven people in the group shakes hands with each of the other six people.
Since the given statement states that there are 42 handshakes, which is twice the actual number, the statement is false.
The correct answer is B) False.